Recommended Posts

Yes, it's for my physics class. I've tried googling this up, and studied the book. But I just DON'T get it.
I understand that pressure is p = p0 + (density of liquid)(gravity)(depth in meters). But when faced with a question, I got stumped. I don't want an answer, I want a clear explanation of why and how it is so.
Okay, the answer is 2440 kg (from back of the book), just to clarify things up.
The question is: A 1.0 m-diameter vat of liquid is 2.0 m deep. The pressure at bottom of the vat is 1.3 atm (or 131,690 Pascals). What is the mass of the liquid in the vat?
Where do I start? I'm stumped. Any help is appreciated! I feel rather dumb, as I know this has to be a simple question.

0

Share this post

Link to post

Share on other sites

The solution to all physics problems is to know all of the equations that include terms that you know, and terms that you need to know, then chain the equations together to get the answer. Basically all you need to memorize are which equations have terms that can help you out and the rest is usually simple algebra (until you get to magnetic fields in calculus-based physics class. SCREW magnetic fields.)

You could probably even make a computer program that does equation lookups for you...HMM

***** SPOILERS BELOW *****

Step 1: Insert all known constants into the pressure equation and solve for density. Your story problem doesn't explicitly state what the pressure at the top of the vat is, but if you think about the name of the units of measure you're using you'll figure it out.

Share this post

Link to post

Share on other sites

It may be useful to look at this from a slightly different POV: you have 0,3 atm extra pressure on the bottom of the vat, having area 0,785. This implies a total force of 23,9 thousand Newtons acting on the said bottom, all due to the liquid sitting above it; so the liquid's weight is 23,9 Newtons, mass is 2430 kg (sorry about the sloppy rounding). Of course, it's the same way one gets the formula you quoted, but sometimes it's easier to handle problems from this direction.